Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.7 AIME, harder Prove it Philippines

Problem:

In $\triangle ABC$ , $AB = 20$ , $BC = 21$ , and $CA = 29$ . Point $M$ is on side $AB$ with $\frac{AM}{MB} = \frac{3}{2}$ , while point $N$ is on side $BC$ with $\frac{CN}{NB} = 2$ . $P$ and $Q$ are points on side $AC$ such that the line $MP$ is parallel to $BC$ and the line $NQ$ is parallel to $AB$ . Suppose that $MP$ and $NQ$ intersect at point $R$ . Find the area of $\triangle PQR$ .

Solution

Solution:

We use similar triangles here. Note that triangles $ABC$ , $AMP$ , $QNC$ and $QRP$ are all similar right (by the Pythagorean theorem, since $20^{2} + 21^{2} = 29^{2}$ ) triangles by AA similarity and corresponding angle theorem. We see that $AP = \frac{3}{5} \cdot 29 = \frac{87}{5}$ and $CQ = \frac{2}{3} \cdot 29 = \frac{58}{3}$ .

Hence, $PQ = AP + QC - AC = \frac{87}{5} + \frac{58}{3} - 29 = \frac{116}{15} = 29\left(\frac{4}{15}\right)$ . Thus, the ratio of similitude between $QRP$ and $ABC$ is $\frac{4}{15}$ , and the area of triangle $QRP$ is $\left(\frac{16}{225}\right)\left(\frac{1}{2}\right)(20)(21) = \frac{224}{15}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.